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### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# White light (consisting of wavelengths from ${\mathbf{400}}{\mathbf{}}{\mathbit{n}}{\mathbit{m}}$ to ${\mathbf{700}}{\mathbf{}}{\mathbit{n}}{\mathbit{m}}$ ) is normally incident on a grating. Show that, no matter what the value of the grating spacing ${\mathbit{d}}$ , the second order and third order overlap.

It is proved no matter what the value of the grating spacing $d$, the second order and third order overlap.

See the step by step solution

## Step 1: Given data

White light has wavelengths between

${\lambda }_{1}=400nm$

${\lambda }_{2}=700nm$

## Step 2: Definition and concept of diffraction from a grating

An optical element with a periodic structure that divides light into a number of beams that move in diverse directions is known as a diffraction grating.

The angular distance of the ${m}_{th}$ order diffraction maxima is given by

${\mathbit{d}}{\mathbit{s}}{\mathbit{i}}{\mathbit{n}}{\mathbit{\theta }}{\mathbf{=}}{\mathbit{m}}{\mathbit{\lambda }}$ …(i)

Here, $\lambda$is the wavelength of the incident light and $d$ is the separation between two lines in the grating.

## Step 3: Determining the angular distances of second and third order maxima

From equation (i) the angular distance of second order maxima from ${\lambda }_{2}$ is

$\begin{array}{l}dsin{\theta }_{2}=2x70\mathrm{mmm}\\ sin{\theta }_{2}=\frac{1400\mathrm{mm}}{d}\end{array}$

From equation (i) the angular distance of third order maxima from ${\lambda }_{1}$ is

$\begin{array}{l}dsin{\theta }_{3}=3x400\mathrm{mm}\\ sin{\theta }_{3}=\frac{1200\mathrm{mm}}{d}\end{array}$

Thus

$\mathrm{sin}{\theta }_{2}>\mathrm{sin}{\theta }_{3}$

Hence the angular distance of a part of the second order from white light is greater than the angular distance of a part of the third order. Thus the second and third order overlap.